Electricity and Magnetism
Review 1: Units 1-6
Mechanics Review 2 , Slide 1
Review Formulas

kq1q2
F1, 2  2 r̂1, 2
r1, 2
Coulomb’s Law
Force law between
point charges

 F
E
q
Electric Field
Force per unit charge
Gauss’ Law
Flux through closed
surface is always
proportional to charge
enclosed
Electric Potential
Potential energy per
unit charge
Electric Potential
Scalar Function that can be
used to determine E
q2
Electric Field
Property of Space
Created by Charges
Superposition
  Qenc
 E  dA 
0
Vab
q1
Gauss’ Law
Can be used to
determine E field
b 

U ab

   E  dl
q
a


E  V
Spheres
Cylinders
Infinite Planes
Applications for Conductors
Conductors
Charges free to move
Field Lines &
Equipotentials
What Determines
How They Move?
They move until
E0!
Spheres
Cylinders
Infinite Planes
E  0 in conductor
determines charge
densities on surfaces
Gauss’
Law
Work Done By E Field
  b  
  F  dl   qE  dl
b
Wab
a
a
Change in Potential Energy
 
U a b  Wa b    qE  dl
b
a
Example
Consider two point charges q1 and q2 located as shown.
1. Find the resultant electric field due to q1 and q2 at the
location of q3.
2. Find the resultant force on q3.
3. Find the electric potential due to q1 and q2 at the
location of q3.
Use components
qi
Ex = å k 2 cosqi
ri
Ey = å k
qi
sin qi
2
ri
q1
q2
V =k
+k
a
2a
Example: Spherical Symmetry
A solid insulating sphere of radius R has uniform
charge density ρ and carries total charge Q. Find the
Electric field everywhere.
y
Choose a suitable Gaussian
surface: A sphere
  Qenc
Qenc
E

d
A


EA


0
0
R
A  4 r 2
x
r
Calculate the charge   Q  Q
V (4 / 3)R 3
enclosed within the
Gaussian surface for r > R
and for r < R
Q
kQ
E


For r > R:
2
2
Qenc  Venc
40 r
Q
r
For r < R:
E
kQ
R
3
r
Example
A solid insulating sphere of radius R has uniform
volume charge density ρ and carries total charge Q.
Find the Potential difference between two points inside
the sphere A and B at distances rA and rB .
rB
y
V  V (rB )  V (rA )    E (r )d r
rA
From the previous problem
we know that for r < R:
E (r ) 
kQ
R
3
R
Q
x
r
V  V (rB )  V (rA )  
kQ
2 R3
(rB2  rA2 )
Example: Cylindrical Symmetry
Find the electric field at a distance r from a line of
positive charge of infinite length and constant
linear charge density λ.
Choose suitable
Gaussian surface:
A cylinder
Calculate the charge
enclosed within the
Gaussian surface
A = 2p rl
Qenc = ll

E
2 r 0
Example: Planar Symmetry
Find the electric field at a distance r due to an
infinite plane of positive charge with uniform
surface charge density σ.
Choose suitable Gaussian
surface: A cylinder
perpendicular to the plane
Calculate the charge
enclosed within the
Gaussian surface
A = pr2
Qenc = s A

E
2 0
Example
Point charge +3Q at center of neutral conducting shell of inner
radius r1 and outer radius r2.
a) What is E everywhere?
Use Gaussian surface = sphere centered on origin
y
  Q
r2
 E  dA 
neutral
conductor
+3Q
r < r1
r1
x
 
2
E

d
A

E
4

r

1 3Q
E
40 r 2
enc
0
r > r2
1 3Q
E
40 r 2
r1 < r < r2
E 0
Example
Point charge +3Q at center of neutral conducting shell of inner
radius r1 and outer radius r2.
a) What is E everywhere?
b) What is charge distribution at r1?
  Qenc
 E  dA   0
y
r2
neutral
conductor
+3Q
r1
r2
x
r1 < r < r2
+3Q
r1
Qenc  0
E 0
1 
 3Q
4 r12
2 
+ 3Q
4 r22
Example
Suppose we give the conductor a charge of Q
a) What is E everywhere?
b) What are charge distributions at r1 and r2?
  Qenc
 E  dA 
0
r < r1
1 3Q
E
40 r 2
r1 < r < r2
E 0
r > r2
1 2Q
E
40 r 2
+ + + + +2Q
+
+ r2
+
+
+3Q
+
r1 +
+
3Q +
+
+ + + +
+
Example
Charge q1 = 2μC is located at the origin. Charge q2 = - 6μC is
located at (0, 3) m. Charge q3 = 3.00 μC is located at (4, 0) m
Find the total energy required to bring these charges to
these locations starting from infinity.
q1q3
q3q2
q1q2
U k
+k
+k
r12
r13
r23
Example
Point charge q at center of concentric conducting spherical
shell of radii a1, and a2. The shell carries charge Q.
What is V as a function of r?
cross-section
Charges q and Q will create an E
field throughout space
1. Spherical symmetry: Use Gauss’
Law to calculate E everywhere
2. Integrate E to get V
 
V (r )    E  d 
r
r0
a2
a1
+Q
+q
metal
Example
  Qenclosed
Gauss’ law:  E  dA 
0
cross-section
a2
a1
+Q
r > a2:
a1 < r < a2 :
+q
metal
r < a2
r
1 Q+q
E (r ) 
4 0 r 2
E (r )  0
1
q
E (r ) 
4 0 r 2
Example
cross-section
a4
a3
+Q
+q
metal
To find V:
1) Choose r0 such that V(r0)  0, usual: r0  ∞
r 

2)Integrate! V ( r )    E  d 
r0
r > a2 :
V (r ) 
Q+q
4  0 r
1
a1 < r < a2 : V (r )  1 Q + q  V (  a2 ) + 0
4  0 a2
r < a1 : V (r )  V (  a2 ) + 0 + V (a1  r )
1 Q+q q q 

V (r ) 
+  
40  a2
r a1 
Example
A rod of length ℓ has a total charge Q uniformly
distributed.
Find V at point P.
dq
V  k
r
dq
dx

r ( x 2 + a 2 )1 / 2
l
V  k  dx
0
(x
1
2
+a
)
2 1/ 2
Example
Charge is uniformly distributed along the x-axis from the
origin to x = a. The charge density is  C/m. What is the xcomponent of the electric field at point P: (x,y) = (a,h)?

dq
E   k 2 rˆ
r
y
P
r
dq   dx
h
x
x
dq
What is 2 ?
r
a
dq
dx

2
r
(a  x) 2 + h 2
Example
Charge is uniformly distributed along the x-axis from the
origin to x = a. The charge density is  C/m. What is the xcomponent of the electric field at point P: (x,y) = (a,h)?

dq
E   k 2 rˆ
r
kdq
kdx
dE  2 
r
( a  x) 2 + h 2
dEx  dE cosq
cosq 
ax
( a  x) 2 + h 2
dE
q
P
y
r
q1
x
dEx
h
q
a
dq   dx
Ex   dEx   dE cosq
x
Example
Charge is uniformly distributed along the x-axis from the
origin to x = a. The charge density is  C/m. What is the xcomponent of the electric field at point P: (x,y) = (a,h)?

dq
E   k 2 rˆ
r
kdq
kdx
dE  2 
r
( a  x) 2 + h 2
dE
y
r
E x   dEx   dE cosq
cosq 
q1
ax
x
( a  x) 2 + h 2
a
E x ( P)  k  dx
0
ax
((a  x)
2
+h
)
2 3/ 2
q
P
dEx
h
q
x
a
dq   dx
k 
h
1 

h 
h2 + a 2



Example
The spheres have the same
known mass m and charge q
and are in equilibrium.
Given the angle θ and the
length L, find the charge q on
each sphere.
Is equilibrium possible if the
charges are different?
Use x and y components
Yes the force will be the
same if the product of
the charges is the same
Example
A solid insulating sphere of radius R has uniform
volume charge density ρ and carries total charge Q.
Find the flux through the outside sphere (r ≥ a). What if
r < R?
S 
Qenclosed
y
0
R
Q
x
r
Qenclosed  Venclosedcharge